written 5.8 years ago by | modified 2.6 years ago by |

Find out the new diameter required

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An old water supply distribution of 250 mm diameter of a city is to replaced by two parallel pipes of equal diameter having equal lengths and identical friction factor values.

written 5.8 years ago by | modified 2.6 years ago by |

Find out the new diameter required

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written 5.6 years ago by |

**Solution**

Dia of old pipe = 'D' = 250mm = 0.25m.

Now, dia. of new pipe which are equal in diameter, length and friction factor value be 'd' (n) each.

**Q =** discharge in old pipe.

**Q1 =** discharge in 1st parallel pipe.

**Q2 =** discharge in 2nd parallel parallel.

**f =** friction factor.

For a single pipe replaced by two parallel pipes, then the head loss will be same in single pipe and in each of parallel pipes i.e.

$h_f = hf_1 = hf_2$

and discharge would be,

$Q = Q_1 + Q_2$

But as per given condition that,

$d_1 = d_2 = d$

and $L_1 =L_2$ of parallel pipes.

$\therefore Q_1 = Q_2$

i.e. $Q = Q_1 + Q_1$

$Q = 2Q_1$

i.e. $Q_1 = Q_2 = Q/2$ -------- (1)

Now, head loss in single pipe i.e. old pipe.

$h_f = \frac{f \times L \times v^2}{D \times 2g}$

$h_f = f \times L \times \frac{ \bigg(\frac{Q}{\pi \times 0.25^2} \bigg)^2}{0.25 \times 2 \times 9.81}$

{ $\therefore v = \frac{Q}{A} = \frac{Q}(\frac{\pi}{4} \times D^2$ }

Now, $hf_1$ = Head loss in 1st parallel pipe.

= $\frac{f \times L_1 \times v_1^2}{d \times 2g}$

$hf_1 = f \times L \times \frac{ \bigg( \frac{Q/2}{\pi \times d^2 } \bigg)^2 }{d_1 \times 2g}$

{ $\because v_1 = \frac{Q1}{A1} = \frac{Q/2}{\pi \times d^2} \because Q_1 = Q/2$ }

= $\frac{f \times L \times (4Q)^2}{d \times 2 \times 9.81 \times ( 2 \times \pi \times d^2)^2}$

But when pipes in parallel

$h_f = hf_1 = hf_2$

$h_f = hf_1$

$\therefore \frac{f \times L \times (4Q)^2}{0.25 \times 2 \times 9.81 \times (\pi \times 0.25)^2}$ = $\frac{f \times L \times (4Q)^2}{d \times 2 \times 9.81 \times (2 \times \pi \times d^2)^2}$

$\therefore d \times (2\pi d^2)^2 = 0.25 \times (\pi \times 0.25^2)^2$

$\therefore d^5 = \frac{0.25^5}{4}$

i.e. d = 0.1894 m

d= 0.19m

Therefore the new diameter required for two pipes in parallel is 0.19m.

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